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aynıdır."]},"beginTime":211,"endTime":338,"href":"/video/preview/12870677740238077564?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=211&ask_summarization=1"},{"index":3,"title":"Analiz Yöntemi","list":{"type":"unordered","items":["Analizde önce koordinat sistemi tanımlanır, kinematik diyagram çizilir ve A ile B noktalarının ivme vektörleri belirlenir.","A ve B noktaları doğrusal hareket yaparsa sıkıntı yoktur, eğrisel hareket yaparlarsa ivmeler teğetsel ve normal bileşenleri ile ifade edilmelidir.","Bağıl ivme denklemi kullanılarak B noktasının ivmesi hesaplanır ve sonuç vektör negatif çıkarsa, kinematik diyagramda çizilen hızların ters yönde olduğu anlaşılır."]},"beginTime":338,"endTime":454,"href":"/video/preview/12870677740238077564?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=338&ask_summarization=1"},{"index":4,"title":"Örnek Problemin Çözümü","list":{"type":"unordered","items":["AB çubuğunda A noktasının anlık ivmesi 5 m/s², anlık hızı 6 m/s olarak verilmiş ve duvara yaslanmış sürtünmesiz bir hareket yapmaktadır.","A noktasının hızı düşey, B noktasının hızı yatay olduğundan, bu iki noktaya dik çizilen doğrultular yardımıyla AB çubuğunun anlık dönme merkezi bulunabilir.","Anlık dönme merkezi etrafındaki açısal hız, A noktasının hızı ve anlık dönme merkezine uzaklığı kullanılarak hesaplanır."]},"beginTime":454,"endTime":635,"href":"/video/preview/12870677740238077564?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=454&ask_summarization=1"},{"index":5,"title":"Bağıl İvme Denkleminin Uygulanması","list":{"type":"unordered","items":["AB = AA + (alfa×rbaa) + (-omega²×rbaa) formülü kullanılarak B noktasının ivmesi hesaplanır.","rbaa vektörünün x bileşeni 3, y bileşeni -4 olarak bulunur ve bu değerler denklemde yerine konulur.","Denklem düzenlenerek i ve j bileşenleri ayrı ayrı eşitlenir ve AB çubuğunun açısal ivmesi ile B noktasının ivmesi hesaplanır."]},"beginTime":635,"endTime":870,"href":"/video/preview/12870677740238077564?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=635&ask_summarization=1"},{"index":6,"title":"Temas Eden Cisimler ve Yuvarlanma Hareketi","list":{"type":"unordered","items":["Temas eden cisimlerde (örneğin dişli) teğetsel ivmeler eşittir ancak normal bileşenleri farklıdır.","Dinamikte karşılaşılan diğer yaygın problem kaymadan yuvarlanma hareketidir.","Kaymadan yuvarlanan top, silindir veya disk gibi cisimlerin hareketi analiz edilebilir ve bu durumda bağıl hız ve bağıl ivme denklemleri yardımcı olur."]},"beginTime":870,"endTime":1022,"href":"/video/preview/12870677740238077564?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=870&ask_summarization=1"},{"index":7,"title":"Yuvarlanma Hareketinde Anlık Dönme Merkezi","list":{"type":"unordered","items":["Yuvarlanma sırasında silindirin merkezi G noktası, yatay düz bir çizgi boyunca ilerler.","Anlık dönme merkezi G noktasında hız sıfırdır.","Yuvarlanma öncesi A noktasının konumu aşağıya doğru hareket ederken, temastan sonra yukarıya doğru yönlenecektir."]},"beginTime":1041,"endTime":1122,"href":"/video/preview/12870677740238077564?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1041&ask_summarization=1"},{"index":8,"title":"Hız ve İvme Hesaplamaları","list":{"type":"unordered","items":["Anlık dönme merkezi G noktasının hızı ω×r veya ω×i birim vektörü şeklinde ifade edilebilir.","G noktasının ivmesi yatay olup, AG ivmesi AG×i şeklinde yazılabilir.","A noktasının ivmesi, zemini terk etme anında yukarı doğru yönlendirmeyi sağlayan bir ivmeye sahiptir ve A×j şeklinde ifade edilebilir."]},"beginTime":1122,"endTime":1297,"href":"/video/preview/12870677740238077564?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1122&ask_summarization=1"},{"index":9,"title":"Örnek Problemler","list":{"type":"unordered","items":["İlk örnek problemde dişli yuvarlanırken, G merkezinin yatay ivmesi α×r = 6×0,3 = 1,80 m/s² olarak hesaplanmıştır.","İkinci örnek problemde AB elemanı saat ibresinin ters yönünde dönüyor ve B noktasının hızı ve ivmesi hesaplanmıştır.","Son örnek problemde C bloğunun veya silindirinin hızı ve ivmesi, bağlı hız ve ivme denklemleri kullanılarak hesaplanmıştır."]},"beginTime":1297,"endTime":2044,"href":"/video/preview/12870677740238077564?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1297&ask_summarization=1"}],"linkTemplate":"/video/preview/12870677740238077564?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"D22 Dinamik - Rijit Cisim Düzlemsel Kinematiği - Bağıl Hareket Analizi - 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İlk olarak tork ve açısal ivme kavramları arasındaki ilişki, Newton'un ikinci yasasıyla benzerliklerle açıklanmakta, ardından katı cisimlerin dönme hareketi için temel denklemler türetilmektedir. Eğitmen, teorik bilgileri vererek silindirin dönme hareketi üzerine iki örnek problem çözmektedir.","Videoda özellikle silindirin etrafına sarılmış kabloyu çekerek oluşan açısal ivme hesaplaması ve düşen bir cismin ivmesi ile kablodaki gerilme kuvvetinin bulunması gibi uygulamalı örnekler incelenmektedir. Silindirin eylemsizlik momenti (1/2 mR²) kullanılarak T = 1/2 m a denklemine indirgeme işlemi gösterilmektedir."]},"endTime":1422,"title":"Dönen Cisimlerin Dinamikleri: Tork ve Açısal İvme","beginTime":0}],"fullResult":[{"index":0,"title":"Tork ve Açısal İvme İlişkisi","list":{"type":"unordered","items":["Bu videoda tork ve açısal ivme kavramları arasındaki ilişki incelenecektir.","Newton'un ikinci yasasında kuvvet ve ivme arasındaki ilişkiyi hatırlayarak, torkla açısal ivme arasındaki benzer bir ilişki kurulacaktır.","Bu ilişki, dönme hareketinin altında yatan temel ilişkiyi anlamamızda önemli olacaktır."]},"beginTime":0,"endTime":40,"href":"/video/preview/7166352186306185824?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=0&ask_summarization=1"},{"index":1,"title":"Dairesel Hareket ve Kuvvetler","list":{"type":"unordered","items":["Sabit z ekseni etrafında dönebilen bir katı cisim düşünülür ve bu cisim küçük noktasal parçacıklardan oluşmaktadır.","Bir parçacık (m₁) z ekseni etrafında dairesel bir rota izler ve bu hareketin yarıçapı r₁ olarak gösterilir.","Parçacığa etki eden dış kuvvet üç bileşenden oluşur: F₁ₜ (radyal bileşen), F₁ₐ (tangansiyel bileşen) ve F₁ₗ (z ekseni doğrultusundaki bileşen)."]},"beginTime":40,"endTime":138,"href":"/video/preview/7166352186306185824?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=40&ask_summarization=1"},{"index":2,"title":"Döndürme Etkisi","list":{"type":"unordered","items":["F₁ₗ ve F₁ₐ kuvvetleri döndürme etkisi oluşturmaz, sadece F₁ₐ (tangansiyel bileşen) döndürme etkisi yaratır.","Newton'un ikinci yasasına göre F₁ₐ = m₁ × a₁ₐ eşitliği yazılabilir, burada a₁ₐ normal ivmedir.","İvme ile açısal ivme arasında r₁ × αₗ = a₁ₐ ilişkisi vardır, burada αₗ açısal 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şeklinde yazılabilir, burada (I₁ + I₂ + ... + Iₙ) cismin eylemsizlik momentidir ve I ile gösterilir.","Sonuç olarak, cisme etki eden net tork Στ = I × αₗ olarak ifade edilir."]},"beginTime":341,"endTime":420,"href":"/video/preview/7166352186306185824?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=341&ask_summarization=1"},{"index":5,"title":"Tork ve Açısal İvme İlişkisinin Önemi","list":{"type":"unordered","items":["Bu eşitlik, torkla açısal ivme arasındaki ilişkiyi tespit etmiştir.","Cisme etki eden net tork, cisimde açısal ivme oluşmasına neden olur.","Newton'un ikinci yasasında kütle, tork ve açısal ivme arasındaki orantı sabiti ise eylemsizlik momentidir."]},"beginTime":420,"endTime":502,"href":"/video/preview/7166352186306185824?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=420&ask_summarization=1"},{"index":6,"title":"İç ve Dış Kuvvetler","list":{"type":"unordered","items":["Cisim oluşturan parçalar birbirine kuvvet etki ettirir ve Newton'un üçüncü yasası gereği etki-tepki çifti olarak düşünülebilir.","İç kuvvetlerin oluşturduğu torklar birbirini götürür, dolayısıyla toplam torka etkisi yoktur.","Cisme etki eden net tork, sadece dış kuvvetlerin yarattığı torklardır."]},"beginTime":502,"endTime":585,"href":"/video/preview/7166352186306185824?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=502&ask_summarization=1"},{"index":7,"title":"Katı ve Sabit Cisimler","list":{"type":"unordered","items":["Katı ve sabit cisimler, dönerken her parçacığı için açısal ivmenin aynı olduğu cisimlerdir.","Sıvılar veya stabil olmayan cisimler için farklı durumlar söz konusudur.","Cismin ağırlığının kütle merkezinde etkisi düşünüldüğünde hesaplamalarda doğru sonuçlar elde edilebilir."]},"beginTime":592,"endTime":704,"href":"/video/preview/7166352186306185824?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=592&ask_summarization=1"},{"index":8,"title":"Tork ve Eylemsizlik Momenti","list":{"type":"unordered","items":["Tork, cisme etki eden net kuvvetin eylemsizlik momenti ile açısal ivmesinin çarpımına eşittir.","Eylemsizlik momenti, cismin kütle ve yarıçapının karesinin çarpımının ikiye bölünmesiyle hesaplanır.","Açısal ivme ile normal ivme arasındaki ilişki, a = rα formülüyle ifade edilir."]},"beginTime":704,"endTime":710,"href":"/video/preview/7166352186306185824?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=704&ask_summarization=1"},{"index":9,"title":"Örnek Problemler","list":{"type":"unordered","items":["Silindirin çevresine sarılmış kabloyu çekerek oluşan açısal ivme ve kablonun ivmesi hesaplanabilir.","Kütleli cismin ivmesi ve kablodaki gerilme kuvveti, Newton'un ikinci yasası kullanılarak bulunabilir.","Silindire etki eden kuvvetler arasında ağırlık, normal kuvvet ve kablonun gerilme kuvveti bulunur."]},"beginTime":710,"endTime":1104,"href":"/video/preview/7166352186306185824?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=710&ask_summarization=1"},{"index":10,"title":"Silindirin Dönmeye İlişkin Eşitlikler","list":{"type":"unordered","items":["Silindirin dönmesini sağlayan net tork, kuvvet çarpı kaldıraç kolu (T×r) ile eylemsizlik momenti çarpı açısal ivme (I×α) arasında bir denklem kurulabilir.","Silindirin eylemsizlik momenti 1/2×m×r² olduğundan, denklem T×r = 1/2×m×r²×α şeklinde yazılabilir.","Düşen cismin ivmesi (a) ile silindirin açısal ivmesi (α) arasında r×α = a ilişkisi vardır."]},"beginTime":1108,"endTime":1291,"href":"/video/preview/7166352186306185824?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1108&ask_summarization=1"},{"index":11,"title":"İvme ve Gerilme Kuvvetinin Hesaplanması","list":{"type":"unordered","items":["Tork denkleminden T = 1/2×m×a ilişkisi bulunur ve bu değer diğer denklemlerde yerine konulabilir.","Düşen cismin ivmesi a = g / (1 + m/2M) olarak hesaplanabilir.","Kabloda (veya ipteki) gerilme kuvveti T = m×g / (1 + 2m/M) olarak bulunur."]},"beginTime":1291,"endTime":1415,"href":"/video/preview/7166352186306185824?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1291&ask_summarization=1"}],"linkTemplate":"/video/preview/7166352186306185824?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"49) Fizik 1 - Dönme hareketinin dinamiği - Tork ve açısal 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Eğitmen, doğrusal hareketin grafiklerini anlatmaktadır.","Video, ivme-zaman grafiklerinin nasıl yorumlanacağı ve bu grafiklerden hız-zaman ve konum-zaman grafiklerine nasıl geçileceği konusunu ele almaktadır. Eğitmen, \"ivme zaman grafiğinden doğrudan hız veya konum bilgisi çıkarılmaz, her zaman hız zaman grafiği çizilerek çözüm bulunmalıdır\" prensibini vurgulamakta ve çeşitli örnek sorular çözerek konuyu açıklamaktadır.","Videoda ayrıca türev kavramı basit örneklerle anlatılarak, konum fonksiyonundan hız ve ivme hesaplamaları gösterilmektedir. İvme-zaman grafiğinin altındaki alanların hız değişimini, hız-zaman grafiğinin altındaki alanların ise konum değişimini gösterdiği görsel olarak açıklanmaktadır."]},"endTime":931,"title":"Doğrusal Hareketin İvme-Zaman Grafikleri","beginTime":0}],"fullResult":[{"index":0,"title":"İvme-Zaman Grafikleri","list":{"type":"unordered","items":["Bu video, doğrusal hareketin grafiklerinin son çalışması olan ivme-zaman grafiklerini ele alıyor.","İvme-zaman grafiklerinde, alanın hızı verdiğini unutmamak önemlidir.","İvme zaman grafiği ile karşılaştığınızda, ivmenin değerini ve ilk hızı bilmeniz gerekir."]},"beginTime":7,"endTime":95,"href":"/video/preview/9998233906047208491?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=7&ask_summarization=1"},{"index":1,"title":"Grafikler Arasındaki İlişki","list":{"type":"unordered","items":["İvme-zaman grafiğinden doğrudan yol hesaplamak mümkün değildir, arada hız zaman grafiği çizilmesi gerekir.","İvme-zaman grafiğinin altındaki alanlar hız değişimlerini verir.","Hız zaman grafiğinin altındaki alanlar ise yolu verir."]},"beginTime":95,"endTime":102,"href":"/video/preview/9998233906047208491?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=95&ask_summarization=1"},{"index":2,"title":"Örnek Sorular","list":{"type":"unordered","items":["İlk örnek soruda, t anında duran bir cismin t aralığında aldığı yollar hesaplanıyor.","İkinci örnek soruda, t anında duran cismin t aralığında x yol aldığı ve 3t'de kaç x alacağı soruluyor.","Üçüncü örnek soruda, t anında hızı olan cismin 2t-3t aralığında kaç x yol aldığı hesaplanıyor."]},"beginTime":102,"endTime":484,"href":"/video/preview/9998233906047208491?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=102&ask_summarization=1"},{"index":3,"title":"İvme-Zaman Grafiği ve Hız 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Eğitmen, öğrencilere fizik problemlerini adım adım çözerek anlatmaktadır.","Videoda toplam on soru çözülmektedir. İlk beş soru eğik düzlem, sürtünme kuvveti, düzgün elektrik alan, esneklik katsayıları ve düzgün çembersel hareket konularını içermektedir. Daha sonra düzgün çembersel hareket, eğik atış hareketi, harmonik hareket, çift yarık deneyi, manyetik alan, öz indüksiyon ve kuvvetler konularındaki sorular çözülmektedir. Son dört soru ise nehirde yüzmeye başlayan bir yüzücü, serbest bırakılan bir topun yere çarpması ve hava ortamındaki bir cismin hız-zaman grafiği konularını kapsamaktadır.","Her soru için eğitmen, gerekli formülleri hatırlatarak, şekilleri çizerek ve hesaplamaları göstererek çözüm sürecini detaylı şekilde açıklamaktadır. Video, fizik sınavlarına hazırlanan öğrenciler için kapsamlı bir kaynak niteliğindedir."]},"endTime":2442,"title":"AYT Fizik Deneme Sınavı Çözüm Videosu","beginTime":0}],"fullResult":[{"index":0,"title":"Eğik Düzlem Üzerindeki Cisimlerin Hareketi","list":{"type":"unordered","items":["Eğik düzlem üzerindeki mv ve 2m kütleli cisimler sürtünmesi ortamda aynı anda serbest bırakıldığında, cisimlerin eğik düzlem üzerindeyken ivmeleri eşittir.","Eğik düzlem üzerinde hareket eden bir cismin ivmesi mg sinüs alfa formülüyle hesaplanır ve kütleden bağımsızdır.","Eğik düzlem üzerinde cisimler aynı süre içerisinde hızlanır, hızları ve yer değiştirmeleri aynı olur, aralarındaki mesafe değişmez."]},"beginTime":38,"endTime":238,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=38&ask_summarization=1"},{"index":1,"title":"Arabanın Hızlanma ve Kayma Durumu","list":{"type":"unordered","items":["Araba hızlanırken ön tekerlek motordan güç alır ve sağa doğru itilir, bu nedenle sürtünme kuvveti hareket yönüyle aynı yönde olur.","Araba kayarak dururken tekerlek dönmüyor, bu nedenle sürtünme kuvveti sol tarafa doğru olur.","Hızlanırken sağ tarafa doğru, kayarak dururken sol tarafa doğru sürtünme kuvveti oluşur."]},"beginTime":238,"endTime":377,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=238&ask_summarization=1"},{"index":2,"title":"Elektrik Alan İçindeki Cisimlerin Hareketi","list":{"type":"unordered","items":["Düzgün elektrik alan içerisinde serbest bırakılan cisimlerin kütleleri eşit olduğunda, elektrik yükleri arasındaki ilişki incelenir.","Cisimler elektrik alan yönünde hareket ederken, aşağı doğru yerçekimi ivmesi ve sağa doğru elektriksel kuvvet etkisiyle hareket ederler.","Sağ tarafa doğru daha fazla öteleme yapan cismin yükü daha fazladır çünkü yükü ivme ile doğru orantılıdır."]},"beginTime":377,"endTime":591,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=377&ask_summarization=1"},{"index":3,"title":"Yayların Enerji ve Kuvvet İlişkisi","list":{"type":"unordered","items":["Esneklik katsayıları sırasıyla k ve 2k olan m ve n yayları normal boyları değiştirmeden P cismine bağlanmıştır.","P cismi sürtünmesi zeminde ok yönünde x kadar kaydırıldığında, m yayını x kadar uzamış, n yayını da x kadar sıkışmış olur.","M ve n yaylarının P cisme uyguladığı kuvvetler eşit değildir, n yayı m'den iki kat fazla kuvvet uygular ve depolanan enerjiler de iki kat fazladır."]},"beginTime":591,"endTime":724,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=591&ask_summarization=1"},{"index":4,"title":"Düzgün Çembersel Hareket Sorusu","list":{"type":"unordered","items":["Bir cisim düşey düzlemde sekiz saniye periyotla düzgün çembersel hareket yaparken, P noktasından geçtikten beş saniye sonra ipten kurtuluyor.","Cismin ipten kurtulduğu anda hızının yatay ve düşey bileşenleri eşittir çünkü açılar 45 derece olduğundan vx ve vy hızları birbirine eşittir.","P noktasından geçtikten altı saniye sonra ipten kurtulmuş olsaydı yatay atış hareketi yapardı, ancak yukarı yönlü düşey atış hareketi yapardı."]},"beginTime":732,"endTime":901,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=732&ask_summarization=1"},{"index":5,"title":"Harmonik Hareket Sorusu","list":{"type":"unordered","items":["Harmonik hareket yapan cismin 40 santim aralıklı K ve L noktalarındaki ivmeleri sırasıyla 3 ve 2 olarak verilmiştir.","Harmonik harekette denge noktasında hız maksimum, kuvvet ve ivme maksimum, uç noktalarda ise hız maksimum, kuvvet ve ivme maksimum olur.","Cismin periyodu 4 saniye olarak hesaplanmıştır."]},"beginTime":901,"endTime":1111,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=901&ask_summarization=1"},{"index":6,"title":"Kayakla Atlama Sorusu","list":{"type":"unordered","items":["Sürtünmeler önemsiz ortamda kayakla atlama yapan sporcu K noktasından ilk hızsız harekete geçerek belirli bir yörüngeyi izler.","Sürtünme yoksa mekanik enerji korunur ve her yerde cismin mekanik enerjisi aynıdır.","H yüksekliği artarsa, L noktasından geçtiği andaki hızı daha fazla olacaktır ve daha yüksek bir noktaya çıkabilir."]},"beginTime":1111,"endTime":1236,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1111&ask_summarization=1"},{"index":7,"title":"Çift Yarık Deneyi","list":{"type":"unordered","items":["Çift yarık deneyinde perde üzerinde ince iki delik (yarık) bulunur ve ışık kaynağı ile ışık ışınları yarıklardan geçerek kırınıma uğrar.","Işık dalgaları yarıklardan geçerken kırınım yaparak perdeye girişim deseni oluşturur ve bu deney ışığın dalga özelliğini kanıtlar.","Saçak aralığı (Δx) formülü: Δx = λL/d'dir, burada λ dalga boyu, L perde ile yarık düzlemi arasındaki mesafe, d ise yarıkların arasındaki mesafedir."]},"beginTime":1239,"endTime":1394,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1239&ask_summarization=1"},{"index":8,"title":"Manyetik Alan ve Yüklü Cisimler","list":{"type":"unordered","items":["Manyetik alan içerisinde fırlatılan yüklü cisimlere hızına dik bir şekilde manyetik kuvvet etki eder ve bu kuvvet qVB formülüyle hesaplanır.","Manyetik kuvvetin yönü dört parmak manyetik alan, baş parmak hız, avuç içi artı yüklü, avuç içi tersi eksi yüklü yönünde olur.","Manyetik alana fırlatılan yüklü parçacığın sürati manyetik alandan dolayı değişmez ve manyetik kuvvet yüklü parçacığın süratini değiştiremez."]},"beginTime":1394,"endTime":1585,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1394&ask_summarization=1"},{"index":9,"title":"Öz İndüksiyon Akımı","list":{"type":"unordered","items":["Öz indüksiyon akımı devrenin akımının değişimine karşı koyar; akım artarken azaltıcı yönde, akım azalırken artırıcı yönde olur.","Öz indüksiyon elektro-motor kuvvetini oluşturan formül: -L(ΔI/Δt)'dir, burada L öz indüksiyon katsayısı (henry), ΔI akım değişimidir, Δt zamandır.","Reostanın sürgüsü eksiye yönünde çekildiğinde direnç azalır, akım artar ve öz indüksiyon akımı devre akımına zıt yönde oluşur."]},"beginTime":1585,"endTime":1745,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1585&ask_summarization=1"},{"index":10,"title":"Dengede Çubuk Problemi","list":{"type":"unordered","items":["Homojen çubuk düşey dengede iken sürtünmesiz düşey duvarın tepkisi N₁, sürtünmeli yatay zeminin tepkisi N₂ ve zemindeki sürtünme kuvveti Fs'dir.","Çubuk bir konumdan iki konumuna getirildiğinde, çubuğun sağ tarafa doğru kayması sonucu N₁ artar ve sürtünme kuvveti de artar.","P (çubuğun ağırlığı) sabit kaldığı için N₂ değeri değişmez, bu nedenle sorunun cevabı N₁ ve Fs artar, N₂ değişmez."]},"beginTime":1745,"endTime":1963,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1745&ask_summarization=1"},{"index":11,"title":"Nehirde Yüzme Problemi","list":{"type":"unordered","items":["Bir nehirde akıntı hızı v olan bir nehirde, suya göre v hızıyla K noktasından yüzmeye başlayan yüzücü, L noktasından karşı kıyıya ulaşıyor.","Yüzücü harekete başladığı andan itibaren düzgün bir şekilde yavaşlayarak L'ye vardığında hızı suya göre v oluyor ve a/v oranı soruluyor.","Yüzücü, akıntı hızı v ile 3d mesafeye ulaşırken, suya göre v hızıyla 4d mesafeye ulaşmak için yavaşlayarak hareket ediyor ve a/v oranı 1 olarak bulunuyor."]},"beginTime":1967,"endTime":2127,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1967&ask_summarization=1"},{"index":12,"title":"Topun Yere Çarpma Problemi","list":{"type":"unordered","items":["H yüksekliğinden serbest bırakılan 2 kg kütleli top 20 metre/saniye hızla yere çarpıyor ve aynı hız büyüklüğünde yukarı doğru hareket ediyor.","Yerin topa uyguladığı itmenin yönü ve büyüklüğü soruluyor.","İtme momentumdaki değişime eşit olduğundan, momentum değişiminin yönü yukarıya doğru 80 m/s² olarak bulunuyor."]},"beginTime":2127,"endTime":2229,"href":"/video/preview/11895540054321160613?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=2127&ask_summarization=1"},{"index":13,"title":"Hava Direnci ve Hız-Zaman Grafiği","list":{"type":"unordered","items":["Hava ortamındaki bir cisme ait hız-zaman grafiği soruluyor.","Düşey doğrultuda cisim hareket ederken, hava direnci k×a×v² formülüyle hesaplanıyor ve hızlandıkça artıyor.","Cisim hızlandıkça sürtünme değeri arttığı için ivme değeri azalıyor ve limit hıza ulaşıncaya kadar eğim azalan bir grafik 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Eğitmen, hasta olduğunu belirterek konuyu anlatmaktadır.","Video, dönerek öteleme ve açısal momentum konusunu kapsamlı şekilde ele almaktadır. İçerik, öteleme hareketi, dönme hareketi ve titreşim hareketi arasındaki farklarla başlayıp, öteleme ve dönme hızlarının hesaplanması, eylemsizlik momenti kavramı, dönme kinetik enerjisi ve eylemsizlik torku konularını içermektedir. Eğitmen, konuyu üç konu anlatım videosu ve iki soru çözüm videosu ile tamamlayacağını belirtmektedir.","Videoda tekerlek, bisiklet tekerlekleri, dişli çarkları gibi günlük hayattan örnekler kullanılarak konular açıklanmakta ve ÖSYM tadında soru çözümleri yapılmaktadır. Ayrıca, ip cambazları, buz pateni yapan kişiler, oklava sallama ve düşme durumunda el açma gibi pratik örneklerle konular somutlaştırılmaktadır. Video, öğrencilere dönerek öteleme hareketinden test ödevi verilmesiyle sonlanmaktadır."]},"endTime":2518,"title":"AYT Fizik Kampı: Dönerek Öteleme ve Açısal Momentum","beginTime":0}],"fullResult":[{"index":0,"title":"Dönerek Öteleme ve Açısal Momentum Konusuna Giriş","list":{"type":"unordered","items":["Vip Fizik YouTube kanalında AYT fizik kampının 57. gününde dönerek öteleme ve açısal momentum konusu anlatılacak.","Konu anlatımı için üç video, soru çözümü için iki video yayınlanacak ve bu konu zor değil, basite yakın bir konudur.","Konu anlatımı AYT fizik kamp kitabından sayfa 233-235 arasında yer alıyor."]},"beginTime":24,"endTime":96,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=24&ask_summarization=1"},{"index":1,"title":"Hareket Türleri","list":{"type":"unordered","items":["Hareket üç ana kategoriye ayrılır: öteleme hareketi, dönme hareketi ve titreşim hareketi.","Öteleme hareketi, bir cismin kütle merkezinin bir noktadan başka bir noktaya belirli bir doğrultuda yer değiştirmesidir.","Öteleme hızı tüm noktalarda aynı yönlü ve eşit büyüklüktedir."]},"beginTime":96,"endTime":190,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=96&ask_summarization=1"},{"index":2,"title":"Dönme Hareketi","list":{"type":"unordered","items":["Bir cismin sabit bir çembersel yörünge etrafında dolanmasına dönme hareketi denir ve bu hareketin hızı dönme hızıdır.","Çember üzerindeki bütün noktaların dönme hızları birbirine eşit değildir çünkü dönme merkezine olan uzaklıkları farklıdır.","Dönme hızı yarıçap ile orantılıdır ve yönü sürekli değişir."]},"beginTime":190,"endTime":384,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=190&ask_summarization=1"},{"index":3,"title":"Dönerek Öteleme Hareketi","list":{"type":"unordered","items":["Dönerek öteleme hareketi, bir cismin hem dönme hareketi hem de öteleme hareketi yapmasıdır.","Cismin hem dönme hızı hem de öteleme hızı vardır ve bu hızların vektörel toplamına anlık hız veya yere göre hız adı verilir.","Tekerleğin dış kabuk için dönme ve öteleme hızları birbirine eşit büyüklüktedir, ancak iç noktalarda bu hızlar farklıdır."]},"beginTime":384,"endTime":496,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=384&ask_summarization=1"},{"index":4,"title":"Öteleme ve Dönme Hızları","list":{"type":"unordered","items":["Dış kabuk için dönme hızı neyse, öteleme hızının da büyüklüğü aynıdır ve bütün noktalarda aynı yönlü ve büyüklüktedir.","Merkez noktasına (O noktası) dönme hızı yoktur çünkü merkeze uzaklığı sıfırdır, ancak öteleme hızı vardır.","Yere göre hız, istenen noktanın tekerleğin değme noktasına uzaklığı ile doğru orantılıdır."]},"beginTime":497,"endTime":642,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=497&ask_summarization=1"},{"index":5,"title":"Bisiklet Örneği","list":{"type":"unordered","items":["Bisikletin öteleme hızı bütün noktalarda aynı yönde ve büyüklüktedir.","Tekerlek sağa doğru dönüyorsa, merkez noktasının dönme hızı sıfırdır, diğer noktaların dönme hızı yarıçapa dik olarak sağa doğrudur.","K noktasının hızı en büyüktür çünkü iki öteleme hızı toplamı vardır, N noktasının hızı v kök, L noktasının hızı ise v'dir."]},"beginTime":642,"endTime":725,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=642&ask_summarization=1"},{"index":6,"title":"Tekerlek Üzerindeki Noktaların Hızları","list":{"type":"unordered","items":["X noktasının yere göre anlık hızı, öteleme hızı ve dönme hızının bileşkesidir.","Y ve T noktalarının yere göre anlık hızları birbirine eşittir çünkü tekerleğin değme noktasına uzaklıkları aynıdır.","Z noktasının yere göre anlık hızı sıfırdır çünkü sağa doğru öteleme ve sola doğru dönme hızları birbirini götürür.","R noktasının yere göre anlık hızı, Y noktasının yere göre anlık hızından büyüktür çünkü R noktasının değme noktasına uzaklığı daha büyüktür."]},"beginTime":725,"endTime":856,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=725&ask_summarization=1"},{"index":7,"title":"Dişli Çarkları Örneği","list":{"type":"unordered","items":["Dişli çarkları eşmerkezli olup yarıçapları sırasıyla 3r ve r'dir.","L dişlisi üzerinde ilerleyen sistemin temas noktası belirlenir ve bu noktaya olan uzaklıklar ile yere göre hızlar doğru orantılıdır.","X noktasının anlık hızı 4r ile orantılıdır (4v), Y noktasının anlık hızı ise 2r ile orantılıdır (2v), bu nedenle hızların oranı 2'dir."]},"beginTime":856,"endTime":1020,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=856&ask_summarization=1"},{"index":8,"title":"Dişlilerin Hızları","list":{"type":"unordered","items":["X noktasının öteleme hızı, öteleme yapan L dişlisinin hızına eşittir.","X noktasının dönme hızı, dönme merkezine uzaklığı 3r olduğundan, L dişlisinin dönme hızının 3 katı olan 3v kadardır.","X noktasının toplam hızı, öteleme hızı (v) ve dönme hızı (3v) toplamı olan 4v'dir, bu da 2v olarak hesaplanır."]},"beginTime":1022,"endTime":1112,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1022&ask_summarization=1"},{"index":9,"title":"Kaymadan Dönen Tekerlek Sorusu","list":{"type":"unordered","items":["Kaymadan dönerek ilerlemek, patinaj çekmeden ilerlemek anlamına gelir ve her döndüğünde ilerler.","Tekerlek bir tur attığında çevresi kadar (2πr) ilerler.","K noktasının tekerleğin hareket ederken izlediği yörünge, hem ileri hem yukarı doğru hareket ederek yayvan bir görüntü oluşturur."]},"beginTime":1112,"endTime":1160,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1112&ask_summarization=1"},{"index":10,"title":"K Noktasının Yörüngesi","list":{"type":"unordered","items":["K noktası tekerlek dönerken geri doğru gitmez, her zaman ileri doğru hareket eder.","Tekerlek yarım tur atınca (180 derece) K noktası yukarı doğru hareket eder, diğer yarım tur atınca aşağı doğru hareket eder.","Tekerlek bir tur attığında K noktası yatayda 6r kadar ilerler, yukarı doğru ise en fazla 2r kadar çıkabilir."]},"beginTime":1160,"endTime":1392,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1160&ask_summarization=1"},{"index":11,"title":"Eylemsizlik ve Eylemsizlik Momenti","list":{"type":"unordered","items":["Eylemsizlik, bir cismin bulunduğu durumu koruma isteğidir ve Newton'un birinci hareket yasası olarak öğrenilmiştir.","Duran veya öteleme hareketi yapan cisimler kütleleri ile, dönme hareketi yapan cisimler ise eylemsizlik momentleri ile etkiye karşı direnç gösterirler.","Dönen cisimlerin dönmeye karşı gösterdikleri dirence eylemsizlik momenti (I) denir ve dönme hareketine karşı gösterilen zorluğa eylemsizlik torku denir."]},"beginTime":1399,"endTime":1449,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1399&ask_summarization=1"},{"index":12,"title":"Eylemsizlik Momentinin Örnekleri","list":{"type":"unordered","items":["Öteleme hareketi yaparken karşılaşılan zorluk kütledir; kütle arttıkça onu hareket ettirmek daha zor olur.","Dönmeye karşı karşılaşılan zorluk eylemsizlik torkudur; I büyük olan cisimleri döndürmek zordur, I küçük olan cisimleri döndürmek kolaydır.","Binalarda dayanıklılığın belirlenebilmesi için kolonların eylemsizlik momentleri hesaplanır."]},"beginTime":1449,"endTime":1532,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1449&ask_summarization=1"},{"index":13,"title":"Eylemsizlik Momentinin Uygulamaları","list":{"type":"unordered","items":["İp cambazları dengede durabilmek için ellerinde uzun bir çubuk taşırlar; çubuğun uzunluğu arttıkça I artar ve dönmeye karşı direnç artar.","Buzun üzerinde kayarken düşme hareketinde refleks olarak elleri açmak, kütleleri dönme noktasından uzaklaştırarak I'yi artırır ve dönmeyi zorlaştırır.","Buz pateni yapan kişiler hızlı dönmek istediklerinde kollarını kapatırlar, yavaşlamak istediklerinde ise kollarını açarak dönme merkezine kütleleri uzaklaştırırlar."]},"beginTime":1532,"endTime":1716,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1532&ask_summarization=1"},{"index":14,"title":"Eylemsizlik Momentinin Özellikleri","list":{"type":"unordered","items":["Eylemsizlik momenti (I) diğer adıyla eylemsizlik torku, birimi kilogram metrekaredir ve m² ile orantılıdır.","Eylemsizlik momenti, cismin kütlesine, kütlenin dönme eksenine olan uzaklığına (yarıçap) ve geometrik şekline bağlıdır.","Dönen cisimlerin kinetik enerjisine dönme kinetik enerjisi denir ve bu enerji hesaplanabilir."]},"beginTime":1716,"endTime":1967,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1716&ask_summarization=1"},{"index":15,"title":"Dönen Cisimlerin Kinetik Enerjisi","list":{"type":"unordered","items":["N tane noktasal kütleden oluşan katı bir cismin dönme kinetik enerjisi, her bir kütlenin kinetik enerjilerinin toplamına eşittir.","Çizgisel hız yerine açısal hız (omega) ve uzaklık (r) bağıntısı kullanıldığında, dönme kinetik enerjisi formülü 1/2 Iω² şeklinde elde edilir.","Dönen cisimlerde, öteleme hareketinde kütle yerine eylemsizlik momenti (I) kullanılır ve hız yerine açısal hız (ω²) kullanılır."]},"beginTime":1969,"endTime":2121,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=1969&ask_summarization=1"},{"index":16,"title":"Dönen ve Öteleyen Cisimlerin Enerjisi","list":{"type":"unordered","items":["Sadece dönme hareketi yapan cisimler (örneğin pervane) sadece dönme kinetik enerjisine sahiptir.","Sadece öteleme hareketi yapan cisimler (örneğin kayarak hareket eden cisim) sadece öteleme kinetik enerjisine sahiptir.","Hem öteleme hem de dönme hareketi yapan cisimler (örneğin tekerlek) hem öteleme hem de dönme kinetik enerjisine sahiptir ve bu nedenle daha fazla enerjiye sahiptir."]},"beginTime":2121,"endTime":2206,"href":"/video/preview/10117074310716210443?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=2121&ask_summarization=1"},{"index":17,"title":"Eylemsizlik Torku","list":{"type":"unordered","items":["Düzgün yapılı türdeş bir çubuğun farklı eksenlere göre eylemsizlik torkları, eksenlere göre kütlelerin uzaklıkları ile orantılıdır.","Çubuğun uç noktasından döndürülmesi en fazla eylemsizlik torku oluşturur, ortasından döndürülmesi ise en az eylemsizlik torku oluşturur.","Düşerken elimizi açmak veya ip cambazı sopasını uzun tutmak, kütleleri dönme merkezinden uzaklaştırarak eylemsizlik torkunu artırarak dönmeye karşı direnç 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Eğitmen, önceki videoda hareket grafikleri hakkında genel bilgiler verdiğini belirterek, bu videoda gerçek bir örnek üzerinden grafik çizimini göstermektedir.","Videoda, hız-zaman grafiğinden konum-zaman ve ivme-zaman grafiklerinin nasıl çizileceği adım adım anlatılmaktadır. Eğitmen önce hız-zaman grafiğinden konum-zaman grafiğini çizmek için alan hesaplama yöntemini göstermekte, ardından hız-zaman grafiğinden eğim alarak ivme-zaman grafiğini çizmektedir. Ayrıca, grafiklerin nasıl yorumlanacağı, yer değiştirme, ortalama hız ve sürat hesaplamaları gibi konular da ele alınmaktadır. Video, grafik sorularında zaman grafiğinden başka bir grafiğe geçiş yapmanın önemini vurgulayarak sonlanmaktadır."]},"endTime":591,"title":"Hareket Grafikleri Çizimi Eğitim Videosu","beginTime":0}],"fullResult":[{"index":0,"title":"Hareket Grafikleri Hakkında Giriş","list":{"type":"unordered","items":["Önceki videoda hareket grafikleri hakkında genel bilgiler verilmiş, bu videoda ise bir örnek üzerinden grafik çizimi gösterilecek.","Hız-zaman grafiğinden konum-zaman ve ivme-zaman grafikleri çizilecek.","Konum-zaman grafiğini çizmek için hız-zaman grafiğinin altında kalan alanın hesaplanması, ivme-zaman grafiğini çizmek için ise hız-zaman grafiğinin eğiminin alınması gerekiyor."]},"beginTime":0,"endTime":65,"href":"/video/preview/17896971472575291405?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=0&ask_summarization=1"},{"index":1,"title":"Konum-Zaman Grafiği Çizimi","list":{"type":"unordered","items":["Hız-zaman grafiği 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arkadaşlar, İntravenöz (IV) damar yolu açma videomuz 2 kısımdan oluşmaktadır.","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/4504605/29098aafba15287b184b321b95375400/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/NbyKfgAAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"19","reqid":"1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=videoid:16696094007078904961","teaser":[{"list":{"type":"unordered","items":["Bu eğitim videosunda anestezi teknikerler Gürcan Çağrı ve Ahmet Helvacı, ayrıca bir sağlık personeli ve yardımcıları Mehmet ile birlikte damar yolu açma (intravenous catheterization) tekniklerini uygulamalı olarak göstermektedir. Videoda Mevlüt adında bir hasta da yer almaktadır.","Video iki ana bölümden oluşmaktadır. İlk bölümde damar yolu açma için kullanılan malzemeler (intraketler, alkol, pamuk, turnike, serum fizyolojik) tanıtılmakta ve intraketlerin renk kodları ile numaralandırılması açıklanmaktadır. İkinci bölümde ise damar yolu açma uygulaması hasta üzerinde gösterilmekte, damar seçimi, turnike kullanımı, sterilizasyon ve damar yolu açma teknikleri adım adım anlatılmaktadır.","Videoda ayrıca 35 derecelik açıyla başlayıp 10-15 derecelik açıya düşürerek damar yoluna erişim, serum fizyoloji uygulaması ve iki tespit uygulaması detaylı olarak gösterilmektedir."]},"endTime":884,"title":"Intravenöz Damar Yolu Açma Eğitim Videosu","beginTime":0}],"fullResult":[{"index":0,"title":"Giriş ve Malzemeler","list":{"type":"unordered","items":["Anestezi teknikerleri Gürcan Çağrı ve Ahmet Helvacı, introvez damar yolu açma uygulamasını iki kısımda anlatacaklar.","Birinci kısımda kullanılan malzemeler tanıtılarak, ikinci kısımda uygulama gösterilecek."]},"beginTime":0,"endTime":28,"href":"/video/preview/16696094007078904961?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=0&ask_summarization=1"},{"index":1,"title":"Kullanılan Malzemeler","list":{"type":"unordered","items":["Damar yolu açmak için introketler, tespit kağıtları, alkol, pamuk, gazlı bez, turnike ve serum fizyolojik kullanılıyor.","Introketler renklerine göre boyutlandırılmış olup, numara büyüdükçe introketler küçülüyor.","Yetişkinlerde genellikle 20 ve 22 numara pembe ve mavi introketler, çocuklarda sarı, yeni doğan bebeklerde mor introketler kullanılıyor."]},"beginTime":28,"endTime":98,"href":"/video/preview/16696094007078904961?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=28&ask_summarization=1"},{"index":2,"title":"Introketlerin Özellikleri","list":{"type":"unordered","items":["Kan verme işlemleri için yeşil introketler, yoğun bakım hastaları için turuncu ve büyük ameliyatlarda gri introketler tercih ediliyor.","İnsanın damarına göre introket seçiliyor, bazı yetişkinlerin damarları çok ince olabilir.","Pembenin 20 numara olduğu bilgisi ezberlenirse, geri kalan numaraları (22, 24, 26) daha kolay hatırlanabilir."]},"beginTime":98,"endTime":326,"href":"/video/preview/16696094007078904961?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=98&ask_summarization=1"},{"index":3,"title":"Malzemelerin Tanıtımı","list":{"type":"unordered","items":["Turnike, damar yolu açılacak yerin 10-15 cm üzerinde bağlanıyor.","Pamuk ve alkol, damar yoluna girecek yerin mikropları kırmak ve sterilize etmek için kullanılıyor.","Damar yolu açıldıktan sonra kanama durumunda kılavuz tel çekilir ve kanama durumunda bezle temizlenir."]},"beginTime":326,"endTime":442,"href":"/video/preview/16696094007078904961?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=326&ask_summarization=1"},{"index":4,"title":"Uygulama Hazırlığı","list":{"type":"unordered","items":["Damar yolu açmadan önce uygun introket, alkol, pamuk, gazlı bez, turnike, tespit bantları ve serum fizyolojik hazırlanıyor.","Yetişkin için bir tepside malzemeler hazırlanabilir, çocuk veya bebek için farklı bir tepsisi oluşturulabilir."]},"beginTime":442,"endTime":504,"href":"/video/preview/16696094007078904961?parent-reqid=1767404888084698-10833282518042973389-balancer-l7leveler-kubr-yp-sas-220-BAL&text=Dimon4Topics&t=442&ask_summarization=1"},{"index":5,"title":"Teorik Uygulama Adımları","list":{"type":"unordered","items":["Damar yolu açılacak bölge kalp seviyesi altında olmalı.","Turnike, damar yolu açılacak bölgenin 10-15 cm üzerinden bağlanır.","Damar yolu muayene edildikten sonra alkolle yukarıdan aşağıya silinir.","35-45 derecelik açıyla damar yoluna girilip, 10-15 derecelik açıya düşürülerek introket iletılır.","Damar yoluna girildikten sonra alkolle kontrol edilir, kılavuz tel çekilip kapak kapatıldıktan sonra tespit 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